A mass of 6 kg is suspended by a rope of length 2 m from the ceilling A force of 50 N in the horizontal direction is applied at the mid - point P of the rope as shown . What is the angle the rope makes with the vertical in equilibrium ? (Take = `10 ms^(-2)`) . Neglect mass of the rope .

Explanation

Figure 5.8(b) and 5.8(c ) are known as free-body diagrams. Figure5.8(b) is the free-body diagram of W and Fig.5.8 (c ) is the free-body diagram of point P .
Consider the equilibrium of the weight aaaaaaaaw.
Clearly , `T_2` = `6 xx 10 `= 60 N.
Consider the equilibrium of the point P under the action of three forces - the tensions `T_1` and `T_2`, and the horizontal force 50 N . The horizontal and vertical components of the resultant force must vanish seperately :
`T_1` cos `theta` = `T_2` = 60 N
`T_1` sin `theta` = 50 N
which gives that
tan `theta` = 5/6 or `theta ` = `tan^(-1)`(5/6) = `40^(@)`
Note the answer does not depend on the length of the rope (assumed massles ) nor on the point at which the horizontal force is applied.